# sum of this simple series [duplicate]

How do you calculate the sum of this series?

$$\sum_{n=1}^\infty 0.5n \left(\frac{2}{3}\right)^n$$

## marked as duplicate by Toby Mak, mrtaurho, Javi, Community♦Apr 30 at 11:06

Note that we can pull out the $$0.5$$ from the summation, and focus on

$$\sum_{n=1}^\infty n \left( \frac 2 3 \right)^n$$

This bears resemblance to a geometric series. We know that, for a geometric series with ratio $$r \in (-1,1)$$,

$$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$

We want to manipulate this expression into the desired form. First, we see the derivative is given by

$$\sum_{n=0}^\infty nr^{n-1} = \frac{1}{(1-r)^2}$$

and multiplication by $$r$$ gives

$$\sum_{n=0}^\infty nr^{n} = \frac{r}{(1-r)^2}$$

Since the first summand of the sum ($$n=0$$) is itself $$0$$, we can start the summation at $$n=1$$ if we choose, which would match the form of the sum we choose to focus on. Take $$r=2/3$$ and we have the "important" sum; once you have the result, multiply by $$1/2$$ again to account for the pulling out of it at the start, and we're done!

• thank you for your answer! My question is, that the derivative(with respect to 'r') of 1/(1-r) is not -1/(1-r)^2? Because (1-r)^(-1) is a composite function. Thank you – TTomi Apr 30 at 10:57
• Gotta remember the chain rule, I keep making this exact mistake. XD If we let $$f(r) = \frac{1}{1-r} = (1-r)^{-1}$$ then $$\frac{df}{dr} = -1\cdot (1-r)^{-2} \cdot \frac{d}{dr} (1-r) = (-1) \cdot (1-r)^{-2} \cdot (-1) = (1-r)^{-2}$$ – Eevee Trainer Apr 30 at 10:59
• thank you, great, you are totally right!! – TTomi Apr 30 at 11:04

A solution withouth calculus

We want to calculate the sum:

$$\sum_{n=1}^{m} n a^n$$

Notice that:

$$\sum_{n=1}^{m} n a^n=1a^1+2a^2+3a^3+...+m a^m =$$ $$=(a^1+a^2+a^3+...+a^m)+(a^2+a^3+...+a^m)+...+(a^{m-1}+a^m)+(a^m)= a^1(1+a^1+...+a^{m-1})+a^2(1+a^1+..+a^{m-2})+...+a^m(1)$$

By geometric series formula this becomes:

$$a^1 \frac{a^{m}-1}{a-1}+a^2\frac{a^{m-1}-1}{a-1}+...+a^m\frac{a^{1}-1}{a-1}=\sum_{n=1}^{m} a^n \frac{a^{m+1-n}-1}{a-1}$$

Now we're almost done:

$$\sum_{n=1}^{m} a^n \frac{a^{m+1-n}-1}{a-1}=\frac{1}{a-1}\sum_{n=1}^{m} (a^{m+1}-a^{n})=\frac{1}{a-1}\left[\sum_{n=1}^{m} a^{m+1}-\sum_{n=1}^{m}a^{n}\right]=\left[ma^{m+1}-\frac{a^{m+1}-1}{a-1}+1\right]=\left[\frac{ma^{m+2}-ma^{m+1}-a^{m+1}+1+a-1}{(a-1)^2}\right]$$

As $$m$$ tend to infinity all the exponential terms tend to $$0$$ since $$|a|<1$$, so it becomes:

$$\frac{a}{(a-1)^2}$$

If $$a=\frac 23$$ this is equal to:

$$6$$

Your sum presents also a factor $$0.5$$ so the final answer is .

$$3$$

:)

• wow, great! Thank you Eureka! – TTomi Apr 30 at 11:00

There is a simple trick, for $$a < 1$$: $$\sum\limits_{n=0}^\infty n a^n = a\partial_a\sum\limits_{n=0}^\infty a^n = a\partial_a \frac{1}{1-a} = \frac{a}{(1-a)^2}$$

• thank you denklo! – TTomi Apr 30 at 10:57

Hint:

$$\color{green}{\sum_{n=1}^\infty na^n}=\sum_{n=0}^\infty(n+1)a^{n+1}=a\color{green}{\sum_{n=1}^\infty na^n}+\sum_{n=0}^\infty a^{n+1}.$$

$$|a|<1\implies(1-a)S=\frac a{1-a}.$$

• Thank you Yves Daoust! – TTomi Apr 30 at 11:07